A geostationary satellite is orbiting the ear | vedclass

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Question

(A) According to Kepler's Third Law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the orbital radius $(r)$: $

Vedclass · Accepted Answer

$6\sqrt{2} 	ext{ hours}$

Vedclass · Answer

(A) According to Kepler's Third Law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the orbital radius $(r)$: $T^2 \propto r^3$.The orbital radius $r$ is the distance from the center of the Earth,given by $r = R + h$,where $h$ is the height above the surface.For the geostationary satellite $(S_1)$: $h_1 = 6R$,so $r_1 = R + 6R = 7R$. The time period $T_1 = 24 	ext{ hours}$.For the second satellite $(S_2)$: $h_2 = 2.5R$,so $r_2 = R + 2.5R = 3.5R$.Using the ratio: $\left(\frac{T_2}{T_1}ight)^2 = \left(\frac{r_2}{r_1}ight)^3$.$\left(\frac{T_2}{24}ight)^2 = \left(\frac{3.5R}{7R}ight)^3 = \left(\frac{1}{2}ight)^3 = \frac{1}{8}$.$\frac{T_2}{24} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}}$.$T_2 = \frac{24}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} 	ext{ hours}$.

$A$ geostationary satellite is orbiting the earth at a height of $6R$ from the earth's surface ($R$ is the earth's radius). What is the period of rotation of another satellite at a height of $2.5R$ from the earth's surface?

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